Why do the subgroups of $Hdg^{2k}(X)$ generated by the cycle classes and Chern classes of vector bundles  coincide in algebraic variety $X$? In Voinsin's book [1], Theorem 11.32 (page 280) says: 
"If X is an algebraic variety, these subgroups of $Hdg^{2k}(X) coincide." 
However, the proof did not show that the subgroup generated by
cycle classes (denoted by $A$) is containded in  the subgroup generated by Chern classes of vector bundles (denoted by $B$) when $k>1$.
In fact it just claims that  $B\subseteq A$.
There are two questions:
(1) How to show $A\subseteq B$?
(2) Why is  the condition that $X$ is an algebraic variety   necessary?  
[1] C. Voisin, Hodge theory and complex algebraic geometry, Vol. I, Cambrige Univ Press, 2002
 A: To complete Donu's answer, the Hodge classes and the group of classes generated by Chern classes of coherent sheaves can be very different even in the Kähler world. Claire Voisin gave an example of a 4-dimensional torus $X$ (hence Kähler) with:
(1) $\mathrm{Hdg}^4(X,\mathbb{Q})\neq0$
(2) $c_2(\mathcal{F})=0$ for any coherent sheave $\mathcal{F}$ on $X$.
The corresponding article is "A counterexample to the Hodge conjecture extended to Kähler varieties" (in IMRN n.20, 2002).
A: To expand slightly on  Minhyong's comment, the key facts can be
found in Fulton's Intersection Theory. If you look  at the comment  following
corollary 18.3.2, you'll see an isomorphism (in slightly different notation)
$$ch:K^0(X)\otimes \mathbb{Q}\cong CH(X)\otimes\mathbb{Q}$$
where $X$ is a nonsingular variety, $K^0(X)$ is the Grothendieck group of
vector bundles, $CH(X)$ is the Chow group of cycles mod rational equivalence,
and $ch$ is the Chern character. After mapping this to rational cohomology, you get exactly
the statement you want.
Note that this is false if


*

*you omit the $\mathbb{Q}$, or

*if you work on a general (compact) complex manifold because there may not be enough vector bundles or subvarieties. To be clear, I mean that conclusion in cohomology is false:
In Zucker, "Hodge conjecture for cubic fourfolds" Compositio 1977, you can find an example 
of a torus with a nonzero integral $(1,1)$ class which is not a divisor, but it would
necessarily lie in the image of $c_1$.

